Forming differential equations from curve families

Solving a DE produces a family of curves (the general solution). The reverse operation is also useful: given a family of curves described by an equation containing $n$ arbitrary constants, find a DE of order $n$ that the family satisfies. This skill is regularly tested on board exams.

The algorithm

Suppose the family is described by $F(x, y, c_1, c_2, \ldots, c_n) = 0$ with $n$ arbitrary constants.

1. Differentiate the equation with respect to $x$. (Treat $y$ as a function of $x$.) Get a new equation.
2. Differentiate again, and again, until you have $n + 1$ equations: the original plus $n$ differentiations.
3. Eliminate all $n$ constants from these $n + 1$ equations.
4. The result is a DE involving only $x$, $y$, $y'$, …, $y^{(n)}$.

The order of the resulting DE is $n$ , same as the number of constants.

Why does it work?

You have $n$ constants in the family. Differentiating once produces one new equation but does not introduce new constants. After $n$ differentiations, you have $n+1$ equations and only $n$ unknown constants , so the constants can (generically) be eliminated.

Worked examples

Example 1. Form the DE for $y = mx + c$ ($m$ and $c$ arbitrary, two constants).

Differentiate: $y' = m$. Differentiate again: $y'' = 0$. This is the DE , no constants present. Order $2$.

Example 2. Form the DE for $y = A e^x + B e^{-x}$.

$y' = Ae^x - B e^{-x}$, $y'' = Ae^x + B e^{-x} = y$. So $\dfrac{d^2y}{dx^2} - y = 0$. Order $2$.

Example 3. Form the DE for $y = A \cos x + B \sin x$.

$y' = -A\sin x + B \cos x$. $y'' = -A \cos x - B \sin x = -y$. DE: $y'' + y = 0$.

Example 4. Form the DE for $y = a e^{2x} + b e^{3x}$.

$y' = 2a e^{2x} + 3b e^{3x}$. $y'' = 4a e^{2x} + 9 b e^{3x}$.

We have three equations in two unknowns $a, b$. Eliminate. From the first two: $a e^{2x} = 3y - y'$, $b e^{3x} = y' - 2y$. (Solve as a 2x2 linear system.) Substitute into the third: $y'' = 4(3y - y') + 9(y' - 2y) = 12y - 4y' + 9y' - 18y = -6y + 5y'$. So $y'' - 5y' + 6y = 0$.

Example 5. Form the DE for the family of all parabolas with axis along $x$-axis and vertex at the origin: $y^2 = 4ax$, $a$ arbitrary (one constant).

Differentiate: $2y y' = 4a$, so $a = \dfrac{yy'}{2}$. Substitute back: $y^2 = 4 \cdot \dfrac{yy'}{2} \cdot x = 2xyy'$. DE: $y^2 = 2xyy'$, or $y = 2xy'$ (assuming $y \neq 0$).

Example 6. Form the DE of all circles passing through the origin with centres on the $y$-axis.

Family: $x^2 + (y - a)^2 = a^2$, i.e. $x^2 + y^2 - 2ay = 0$. One constant: order $1$.

Differentiate: $2x + 2yy' - 2a y' = 0$, so $a = \dfrac{x + yy'}{y'}$ (assuming $y' \neq 0$). Substitute back: $x^2 + y^2 = 2y \cdot \dfrac{x + yy'}{y'}$, giving $(x^2 + y^2) y' = 2y(x + yy') = 2xy + 2y^2 y'$. Rearrange: $(x^2 + y^2 - 2y^2) y' = 2xy$, so $(x^2 - y^2) y' = 2xy$, i.e. $\dfrac{dy}{dx} = \dfrac{2xy}{x^2 - y^2}$.

A note on the geometric setup

When you read "family of all parabolas with vertex at origin and axis along $x$-axis," translate carefully:

• Family: a parametrised collection of curves.
• Parameters: count them. They become the constants.
• Equation form: write the most general expression that fits the description.

The number of parameters equals the order of the resulting DE.

Try it yourself

1. Form the DE for $y = e^x(A \cos x + B \sin x)$.
2. Form the DE for $x^2 + y^2 = a^2$ (circles centred at origin).
3. Form the DE for the family of all lines through the origin.
4. Form the DE for $y = a x^3$.
5. Form the DE for $xy = c$ (rectangular hyperbolas).
6. Form the DE for $y = c_1 + c_2 e^x$.
7. Form the DE for $y = e^{ax}\sin(bx)$ ($a$, $b$ arbitrary, two constants).
8. Form the DE for $y = a x^2 + bx + c$ ($a$, $b$, $c$ arbitrary).
9. Form the DE of all parabolas with axis parallel to $y$-axis.
10. Form the DE of all circles passing through the origin with centres on the $x$-axis.
11. Form the DE of $(x - a)^2 + (y - b)^2 = r^2$ , all circles of fixed radius $r$, with $a, b$ arbitrary.
12. Form the DE for $y = a \log x + b$.
13. Form the DE for the family of curves $y = \dfrac{A}{x} + B$.
14. Form the DE for the family of ellipses $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ ($a$, $b$ arbitrary).

Pitfalls and tricks

• Count constants carefully. Number of constants $=$ order of resulting DE.
• Eliminate, don't solve. You don't need explicit values for the constants , just remove them.
• Use algebraic substitution to eliminate. From the differentiated equations, solve for each constant as a function of $x$, $y$, $y'$, …, then substitute.
• The final answer should contain only $x$, $y$, $y'$, $y''$, …, and constants from the problem (no arbitrary parameters).
• Recognise hidden constants. $\sin(ax + b)$ has two constants; $\sin(ax) + b$ also has two but different DEs.